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(C) The initial energy of the system is the kinetic energy of the electron, $E_i = 3 \, eV$.The energy of the hydrogen atom in the $n^{th}$ state is g

Vedclass · Accepted Answer

$1.41$

Vedclass · Answer

(C) The initial energy of the system is the kinetic energy of the electron, $E_i = 3 \, eV$.The energy of the hydrogen atom in the $n^{th}$ state is given by $E_n = -\frac{13.6 \, eV}{n^2}$.The second excited state corresponds to $n = 3$. Thus, $E_f = -\frac{13.6 \, eV}{3^2} = -\frac{13.6}{9} \, eV \approx -1.51 \, eV$.The energy released as a photon is the difference between the initial and final energy: $E_{photon} = E_i - E_f = 3 \, eV - (-1.51 \, eV) = 4.51 \, eV$.The work function $\phi$ of the metal is $\phi = \frac{hc}{\lambda_{threshold}} = \frac{12400 \, eV \cdot Å}{4000 \, Å} = 3.1 \, eV$.Using Einstein's photoelectric equation, the maximum kinetic energy is $KE_{max} = E_{photon} - \phi = 4.51 \, eV - 3.1 \, eV = 1.41 \, eV$.

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